As well as time suppose two object of mass are respectively 1 kg and 10 kg released from a building top height 60 m .
under free fall motion (air resistance neglected ) can you tell me which one will appear on the ground first.
Well it’s simple acceleration does not depend upon mass and height for both the object is same so time taken to fall on the ground must be same.
For both object because it depends upon time and distance from earth.
Now this is acceleration hence all kinematics formula will be applicable here.
But motion must be rectilinear (straight line motion only) when an object free fall under gravity.
this motion is called straight line or rectilinear motion we will use acceleration here g.
Here for sign of vector we consider upward direction positive and downward direction negative .
It is arbitrary we can choose opposite also but generally in our Cartesian coordinate system upward direction is taken as positive.
And downward is taken as negative but remember g direction is always downward because earth always pull towards its center.
now apply rectilinear motion equation.
Equation of motion under gravity
v = u -gt (here velocity of object is in upward direction so it is taken as positive and g is in downward direction taken as negative).
h = ut – 1/2gt² (here h is used as height )
v² = u² -2gh
v = -u -gt
h = -ut – 1/2gt²
v² = u² +2gh (here g and h are are negative )
now we will see some basic question on this equation.
Q A stone is dropped from a top of a cliff and is found to travel 44.1 m in the last second before it reaches the ground find height of the cliff.
Ans read question carefully and visualize in your mind before solving any numerical question and note down given parameter.
and the parameter to find then next step find the relation between given parameter and those parameter to be find.
here it is clear that anything drop and start means its initial velocity is zero u = 0 like a ball drop, a car start means u = 0 no body will tell, you have to understand this word .
again in last one second distance travel is 44.1 m given
let total height of cliff be h .
now we have study the equation distance travel in nth second remember Snth = u +1/2a(2n -1) here given Snth = 44.1 m , u = 0 put and find value of n
44.1 = 0+1/2*9.8(2n-1) or 44.1*2 = 2n*9.8 – 9.8 0r n = 5 here total time taken is 5 s
now use equation in 5 s distance travel will be height of cliff
h = -u -1/2gt² = 0 – 1/2*9.8*5*5 h = 122.5 m.
Question for motion under gravity
Q A stone is dropped from a baloon at an altitude of 300 m how long the stone will take to reach the ground if baloon is ascending with a velocity of 5 m/s , if descending with velocity of 5 m/s , if stationary
Ans here read question carefully what is given first
h = 300 m, a = g = 9.8 m/s² downward
important point when two body are moving together and whenever we release one body the velocity of release body will be same whatever speed was together then after gravity will acts upon it then here what will be initial velocity of stone same as velocity of baloon in case of ascending upward with 5 m/s now
h = -300 m ( here displacement is downward )
g = -9.8 approx g = 10
u = 5 m/s
then use equation h = ut+1/2at² or -300 = 5t -1/2*10t²
after solving this equation 5t² – 5t -300 =0 or t² -t – 60 = 0 solve this quadratic equation you will get value of t now in second case when baloon is descending what will be u here just u = -5 m/s other parameter same now in third case when baloon is stationary then u = 0 other parameter will be same put value and get value of time .